Answer :
To determine what happens to the wavelength when the frequency increases, let's examine the relationship between the two.
The relationship between wavelength ([tex]\(\lambda\)[/tex]) and frequency ([tex]\(f\)[/tex]) of a wave is governed by the equation of the speed of light ([tex]\(c\)[/tex]):
[tex]\[ c = \lambda \times f \][/tex]
Here:
- [tex]\( c \)[/tex] is the speed of light in a vacuum, which is a constant approximately equal to [tex]\( 3 \times 10^8 \)[/tex] meters per second.
- [tex]\( \lambda \)[/tex] is the wavelength.
- [tex]\( f \)[/tex] is the frequency.
We can rearrange this equation to solve for the wavelength:
[tex]\[ \lambda = \frac{c}{f} \][/tex]
From this equation, it is clear that the wavelength ([tex]\(\lambda\)[/tex]) is inversely proportional to the frequency ([tex]\(f\)[/tex]). This means that if the frequency increases, the wavelength must decrease to keep the speed of light constant.
Therefore, when the frequency increases, the wavelength decreases.
So, the correct answer is:
O The wavelength decreases.
The relationship between wavelength ([tex]\(\lambda\)[/tex]) and frequency ([tex]\(f\)[/tex]) of a wave is governed by the equation of the speed of light ([tex]\(c\)[/tex]):
[tex]\[ c = \lambda \times f \][/tex]
Here:
- [tex]\( c \)[/tex] is the speed of light in a vacuum, which is a constant approximately equal to [tex]\( 3 \times 10^8 \)[/tex] meters per second.
- [tex]\( \lambda \)[/tex] is the wavelength.
- [tex]\( f \)[/tex] is the frequency.
We can rearrange this equation to solve for the wavelength:
[tex]\[ \lambda = \frac{c}{f} \][/tex]
From this equation, it is clear that the wavelength ([tex]\(\lambda\)[/tex]) is inversely proportional to the frequency ([tex]\(f\)[/tex]). This means that if the frequency increases, the wavelength must decrease to keep the speed of light constant.
Therefore, when the frequency increases, the wavelength decreases.
So, the correct answer is:
O The wavelength decreases.
